Heat absorbed by the water + heat absorbed by the vessel + heat given up by the heated copper tubing = 0
[mass H2O x specific heat water x (Tfinal-Tinitial)] + [10 J/C*(Tfinal-Tinitial)] + mass copper x specific heat copper x (Tfinal-Tinitial) = 0
Solve for Tfinal. Post your work if you get stuck.
A 522g piece of copper tubing is heated to 89.5C and placed in an insulated vessel containing 159g of water at 22.8C. Assuming no loss of water and a heat capacity for the vessel of 10.0J/K, what is the final temperature of the system ( c of copper = 0.387 J/g*K)?
4 answers
89
36.6
Oops I did 455 g Cu to get my answer.